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Basic trigonometry in dynamics force problem

  1. Apr 30, 2017 #1
    1. The problem statement, all variables and given/known data
    Two spheres, with [itex]0.5g[/itex] each, are hanging by [itex]30cm[/itex] threads, tied on the same spot. The same electric charge is communicated to each sphere; in consequence, the threads move apart until they are about [itex]60^\circ[/itex] from each other. What is the value of the charge?

    [itex]\theta = \frac{60^\circ}{2} = 30^\circ \\
    \\m = 0.5g = 0.0005kg\\ d = 30cm = 0.3m[/itex]


    2. Relevant equations
    Not needed.

    3. The attempt at a solution
    So, what I did was the following:

    [tex]\tan \theta = \frac{\vec F}{\vec P} \\ \Rightarrow \frac{\sqrt 3}{3} = \frac{\vec F}{m\vec g} = \frac{\vec F}{0.0005 \times 9.8} = \frac{\vec F}{0.005} \\ \Rightarrow \vec F = \frac{\sqrt 3}{3} \times 0.005 = 0.0028 \\[/tex][tex]k \frac{q^2}{d^2} = 0.0028 \\ \Rightarrow 9 \times 10^9 \times \frac{q^2}{0.3^2} = 2.8 \times 10^{-3} \\ \Rightarrow q^2 = \frac{(2.8 \times 10^{-3}) \times (9 \times 10^{-2})}{9 \times 10^9} = 2.8 \times 10^{-14} \\ \Rightarrow q = 1.7 \times 10^{-7}C[/tex]My result did match the answer in the textbook, but searching through the web, I came accross a different way of solving it:

    v6h0Lxr.png
    The rest is just like mine, so I didn't bother putting it. Anyway, he reached the same result, but with a different method and trigonometric relations, which I'm quite lacking in and I couldn't understand very well yet.

    Why did he conclude that
    ca416d5dfd214713807a96b6c4bfef7d.png

    and
    56f040f51842463aa3cf040d13f2222e.png
    ?

    I know I could just go on and ignore this because I already solved the exercise, but vectors are a puzzling topic for me and I feel like there's something I'm missing... I'd appreciate some help!
     
  2. jcsd
  3. Apr 30, 2017 #2
    i hope that help
    prove.jpg
     
  4. Apr 30, 2017 #3
    the d part :
    prove 2.jpg
     
  5. Apr 30, 2017 #4
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